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Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1@apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1okay, expand the two wholesquares, it's simple enough...

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1the options are 4cos^2(ab/2) 4 sin a/2 cos b/2

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1@apoorvk after expanding i got this then next..............

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.1great, now.. cosAcosB + sinAsinB = cos(AB) Use this^

patmyer1
 2 years ago
Best ResponseYou've already chosen the best response.0ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them

AndrewNJ
 2 years ago
Best ResponseYou've already chosen the best response.1spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(ab)

AndrewNJ
 2 years ago
Best ResponseYou've already chosen the best response.1(cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(ab)

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1ok @apoorvk so it is not over yet!!@

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.2Continuing from where @AndrewNJ left, \[2(1+\cos(ab))\] now we know that \[1 + \cos 2a = 2 \cos^2 a\] Now try :D

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1@shivam_bhalla did nt understand!

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.2= \[2(1+\cos 2\frac{(ab)}{2})\] Now we have a identity, \[1 + \cos 2x = 2 \cos^2 x\] in your case, \[x = \frac{ab}{2}\] Now try

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1i got it thxx @shivam_bhalla

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.1@shivam_bhalla plzz help http://openstudy.com/study?signup#/updates/4fd60c0be4b04bec7f170b7d
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